Question

A mixture of ethane and ethene occupied 41 L at 1 atm and 500 K. The mixture reacts completely with
10
3 mol of O2 to produce CO2 and H2O. The number of moles of ethane in the mixture is

Answer 1

Answer:

subsdh sheb hi nosho bhki

Answer 2

Explanation:

The mole fraction of ethane is 0.667 and ethene is 0.334 in the mixture.

Explanation:

Given data:  

Pressure, P = 1 atm

Volume, V = 41 L

Absolute Temperature, T = 500 K

To find: mole fraction of ethane and ethene in the mixture

 

According to the ideal gas law, we have

PV = nRT

Where R is the ideal gas constant = 0.082 L atm/K mol and n is the no of moles of gas

∴ 1 * 41 = n * 0.082 * 500

Or, n = 1

Therefore, the total moles of (ethane + ethene) = 1

Let the no. of moles of ethane ne “n1” and ethene be “n2”.  

C2H6 + C2H4 = 1

∴ n1 + n2 = 1 ……(i)

We are given that ethane and ethene reacts completely with 10/3 moles of O2 to produce CO2 and H2O.

C2H6 + 7/2 O2 → 2CO2 + 3H2O

C2H4 + 3 O2 → 2CO2 + 2H2O

From these above chemical equation we can write as

7/2n1 + 3n2 = 10/3 …. (ii)

 

Subtracting eq. (i) from (ii), we get

n1 = 2/3 = 0.667

Substituting value of n1 in eq. (i), we get

2/3 + n2 = 1

Or, n2 = 1 - 2/3 = 1/3 = 0.334

Mole fraction of ethane = no. of mole of ethane / total moles = 0.667/1 = 0.667

Mole fraction of ethene = no. of mole of ethene / total moles = 0.334/1 = 0.334